Mid Term Review

Terry A. Ring

CH EN 5253

Design II

Week Day Lecture Topic AssignmentsDue Date

1 12-JanCourse Overview Lecture Review Chapters 7,8,9

14-JanReview of Project EconomicsGeneration of Economics Spread Sheet HW 1

16-JanReview of Project Economics Attainable Rection - HW 2

2 19-JanMLK Holiday

21-JanReview of Reactors - Selectivity & Heat Effects Separation Trains, HW 2 Assigned HW 1

23-JanAttainable Region

3 26-JanSeparations

28-JanDistillation Trains HW-3 Assigned HW 2

30-JanSeparations and Reactors

4 2-FebReactor, Separation and Recycle

4-FebReactor, Separation and Recycle HW-4 Assigned HW 3

6-FebReactor, Separation and Recycle

5 9-FebHeat and Power Integration

11-FebHeat and Power Integration HW-5 Assigned HW 4

13-FebHeat and Power Integration

6 16-FebPresident's Day Holiday

18-FebOptimization on Process Flowsheets HW-6 Assigned HW 5

20-FebEffects of Impurities on Reactors - HX

7 23-FebEffects of Impurites on Separators, Recycle

25-FebPlantwide Control HW-7 Assigned HW 6

27-FebPlantwide Control

8 2-MarReview for Exam

4-MarSequential Batch Processing HW 7

6-MarMid Term Exam Exam

Reactor Heat Effects

S,S&L Chapter 7

Managing Heat Effects

• Reaction Run Away– Exothermic

• Reaction Dies– Endothermic

• Preventing Explosions

• Preventing Stalling

Temperature Effects

• Thermodynamics/Equilibrium

• Kinetics

Unfavorable Equilibrium

• Increasing Temperature Increases the Rate

• Equilibrium Limits Conversion

Reactor with Heating or Cooling

Q = UA ΔT

Best Temperature Path

Optimum Inlet TemperatureExothermic Rxn

Inter-stage Cooler

Exothermic Equilibria

Lowers Temp.

Inter-stage Cold Feed

Exothermic Equilibria

Lowers TempLowers Conversion

Reaction Selectivity

• Parallel Reactions– A+BR (desired)– AS

• Series Reactions– ABC(desired)D

• Independent Reactions– AB (desired)– CD+E

• Series Parallel Reactions– A+BC+D– A+CE(desired)

• Mixing, Temperature and Pressure Effects

Rate Selectivity

• Parallel Reactions– A+BR (desired)– A+BS

• Rate Selectivity

• (αD- αU) >1 make CA as large as possible• (βD –βU)>1 make CB as large as possible

• (kD/kU)= (koD/koU)exp[-(EA-D-EA-U)/(RT)]– EA-D > EA-U T– EA-D < EA-U T

)()(A

U

Drr

D/UD

U

D Ck

kS UDU

BC

Reactor Design to Maximize Desired Product

Maximize Desired Product

• Series Reactions– AB(desired)CD

• Plug Flow Reactor• Optimum Time in Reactor

Real Reaction Systems

• More complicated than either – Series Reactions– Parallel Reactions

• Effects of equilibrium must be considered

• Confounding heat effects

• All have Reactor Design Implications

Engineering Tricks

• Reactor types– Multiple Reactors

• Mixtures of Reactors

– Bypass– Recycle after Separation

• Split Feed Points/ Multiple Feed Points• Diluents• Temperature Management

Sorted Out with Attainable Region Analysis

Attainable Region

S,S&L Chapt. 7

Attainable Region

• Graphical method that is used to determine the entire space feasible concentrations

• Useful for identifying reactor configurations that will yield the optimal products

ProcedureStep 1: Construct a trajectory for a PFR from the feed point,

continuing to complete conversion or chemical equilibriumStep 2: When the PFR bounds a convex region, this constitutes a

candidate AR. The procedure terminates if the rate vectors outside the candidate AR do not point back into it.

Step 3: The PFR trajectory is expanded by linear arcs, representing mixing between the PFR effluent and the feed stream, extending the candidate AR.

Step 4: Construct a CSTR trajectory to see if the AR can be extended. Place linear arcs, which represent mixing, on the CSTR trajectory to ensure the trajectory remains convex.

Step 5: A PFR trajectory is drawn from the position where the mixing line meets the CSTR trajectory. If the PFR trajectory is convex, it extends the previous AR to form a expanded AR. Then return to step 2. Otherwise, repeat the procedure from Step 3.

Example

BBAB

ABAA

k

kk

k

CkCkCkdt

dC

CkCkCkdt

dC

DA

CdesiredBA

321

2421

4

31

2

2

)(

Reactions

Rate Equations

Step 1

Begin by constructing a trajectory for a PFR from the feed point, continuing to the complete conversion of A or chemical equilibrium

• Solve the PFR design equations numerically– Use the feed conditions as initial conditions to

the o.d.e.– Adjust integration range, (residence time),

until complete conversion or to equilibrium

PFR Design Equations

BBAB

ABAA

CkCkCkdt

dC

CkCkCkdt

dC

321

2421

x

AA r

dxFV

0

0

Solve NumericallyRunge-Kutta

Solve Numerically

Step 2Plot the PFR trajectory from the previous results. Check to see if rate vectors outside AR point back into it (e.g. Look for non-convex regions on the curve. Tangent line passing (1,0))

Des

ired

Step 3

Expand the AR as much as possible with straight arcs that represent mixing of reactor effluent and feed stream

PFR

(1-)

Interpreting points on mixing line

Larger Attainable Region

PFR CA=0. 2187CB=0.00011 CA=0.72

CB=0.00004

CA=1CB=0

(1-)PFR

CA=1CB=0

(1-)

Mixing of StreamsReactant Bypass

21 )1( ccc Vector Equation, i component is CA, j component is CB

α =fraction of mixture of stream 1in the mixed stream

)1(00011.0000004.0

)1(2187.0172.0

B

A

C

C

Feed mixing fraction: = 0. 64

Step 4

If a mixing arc extends the attainable region on a PFR trajectory, check to see if a CSTR trajectory can extend the attainable region

For CSTR trajectories that extend the attainable region, add mixing arcs to concave regions to ensure the attainable region remains convex

• Solve CSTR multiple NLE numerically– Vary until all feed is consumed or equilibrium is

reached

CSTR Design Equations

)(

)(

321

2421

BBAB

ABAAAo

CkCkCkC

CkCkCkCC

A

A

r

xFV

0

Solve numerically at various until complete conversion or

equilibrium is achieved

CSTR Extends Attainable Region

CSTR

Plot extends attainable regioni.c. for step 5

CSTR

CA=1CB=0

(1-)

EnlargesAttainableRegion

Possible Configuration at this point

CSTR

CA=1CB=0

βPFR

β = 0

β = 0

1-α-β

0.38

β = 1α = 0

Profit ($) = 15000*CB-15*CA2

Attainable Region

-0.00001

0.00001

0.00003

0.00005

0.00007

0.00009

0.00011

0.00013

0.00015

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 1.05

C a, kmol / m3

CSTR

PFR

PFR2

$=0.9

$=2

$=1.5

Optimal point not at highest selectivity

PFRCSTR

Conclusions

• Need to know feed conditions• AR graphical method is 2-D and limited to 2

independent species• Systems with rate expressions involving more

than 2 species need to be reduced– Atom balances are used to reduce independent

species– Independent species = #molecular species - #atomic

species• If independent species < 2, AR can be used by Principle of

Reaction Invariants

Separation Trains

S, S&L Chapt. 8

Separation Methods

• Absorption

• Stripping

• Distillation

• Membrane Separations

• Crystallization

• etc

Use of Separation Units

Column Sequences

• No. of Columns– Nc=P-1

• P= No. of Products

• No. of Possible Column Sequences– Ns=[2(P-1)]!/[P!(P-1)!]

• P= No. of Products

– P=3, Nc=2, Ns=2– P=4, Nc=3, Ns=5 – P=5, Nc=4, Ns=14– P=6, Nc=5, Ns=42– P=7, Nc=6, Ns=132

No. of Possible Column Sequences Blows up!

How do I evaluate which is best sequence?

Marginal Vapor Rate

• Marginal Annualized Cost~ Marginal Vapor Rate• Marginal Annualized Cost proportional to

– Reboiler Duty (Operating Cost)– Reboiler Area (Capital Cost)– Condenser Duty (Operating Cost)– Condenser Area (Capital Cost)– Diameter of Column (Capital Cost)

• Vapor Rate is proportional to all of the above

Selecting Multiple Column Separation Trains

• Minimum Cost for Separation Train will occur when you have a– Minimum of Total Vapor Flow Rate for all

columns– R= 1.2 Rmin – V=D (R+1)

• V= Vapor Flow Rate• D= Distillate Flow Rate• R=Recycle Ratio

Azeotrope Conditions

• Conditions on the Activity Coefficient

• Minimum Boiling, γjL> 1

• Maximum Boiling, γjL< 1

• xj=yj, j=,1,2,…C

Law sRaoult' from Deviations Negative,...2,1,1

Law sRaoult' from Deviations Positive,...2,1,1

)1( 221111

Cj

Cj

PxPxP

Lj

Lj

sLsLT

Raoult’s Law

satii PxP

Importance of Physical Property Data Set

• In all cases– Need sophisticated liquid phase model to

accurately predict the activity coefficient for the liquid.

• For High Pressure Cases Only– Also need sophisticated (non-ideal) gas

phase fugacity model

Multi-component Azeotropes

• Residue Curve Map– dxj /dť = dxj /d ln(L) = xj – yj

• Integrate from various starting points

Defining Conditions for Multi-component Azeotrope

t goes from 0 to 1, ideal to non-ideal to find Azeotrope

Distillation

• XB, XF and YD form a line for a Distillation Column

• Line can not cross Azeotrope line

Ethanol/Water Distillation with

BenzeneTo Break Azeotrope

Pressure Swing to Break Azeotrope

Temp. of Azeotropevs. Pressure

Mole Fraction of Azeotrope

Reactor-Separation Train-Recycle

Chapt. 7&8

Trade-off between Reactor and Separator

• Factors– Reactor Conversion of limiting reactant

• Effects cost and size of Separation Train

– Reactor Temperature and mode of operation (adiabatic, isothermal, etc.)

• Effect utility costs for separation and reaction• Effect impurities from side reactions

– High Reactor Pressure for Le Chatlier cases (less moles of product)

• Higher cost for recycle compression

Trade-off between Reactor and Separator

• Factors, cont.– Use of excess of one or more reactant to increase

equilibrium conversion and/or reaction rate• Increases cost of separation train

– Use of diluents in adiabatic reactor to control temperature in reactor

• Increases cost of separations train

– Use of purge to avoid difficult separation.• Decreases the cost of separations• Loss of reactants – increase cost of reactants• May increased cost of reactor, depending on the purge-to-

recycle ratio

Factors that effect recycle/purge

• Factor– Excess reactants

• Increases recycle flow• Increases separation costs

– Concentration of impurities to be purged• Effects the recycle-to-purge ratio

– Reactor outlet temperature and pressure• Increase cost of utilities in separation• Increase cost to recycle - compressor

Compare Recycle Concepts

• Costs

• Benefits

Feedback effects of Recycle Loop

• Small disturbance on feed

• Large effect on recycle flow rate/composition

• Snowball effect on reactor/separator

Heat Integration

Chapter 9

Terry Ring

University of Utah

Costs

• Heat Exchanger Purchase Cost– CP=K(Area)0.6

• Annual Cost– CA=im[ΣCp,i+ ΣCP,A,j]+sFs+(cw)Fcw

• im=return on investment• Fs= Annual Flow of Steam,

– $5.5/ston to $12.1/ston = s

• Fcw=Annual Flow of Cold Water– $0.013/ston = cw

Lost Work = Lost Money

• Transfer Heat from T1 to T2

• ΔT approach Temp. for Heat Exchanger

• To= Temperature of Environment

• Use 1st and 2nd laws of Thermodynamics

• LW=QToΔT/(T1T2)– ΔT=T1-T2

– To= Environment Temperature

• Q= UAΔTlm

T1

T2

Q

Heat Integration

• Make list of HX• Instead of using utilities can you use

another stream to heat/cool any streams?• How much of this can you do without

causing operational problems?• Can you use air to cool?

– Air is a low cost coolant.

• Less utilities = smaller cost of operations

Terms

• HEN=Heat Exchanger Network

• MER=Maximum Energy Recovery

• Minimum Number of Heat Exchangers

• Threshold Approach Temperature

• Optimum Approach Temperature

Process

Minimize UtilitiesFor 4 Streams

Adjust Hot Stream Temperatures to Give ΔTmin

Enthalpy Differences for Temperature Intervals

Pinch Analysis

Minimum Utilities

Pinch Analysis

ΔTapp

MER values

How to combine hot with cold?

• At Pinch (temp touching pinch)– Above Pinch Connect

• Cc≥Ch

– Below Pinch Connect• Ch≥Cc

• Not touching Pinch temp.– No requirement for Cc or Ch

4 Heat ExchangerHEN for Min. Utilities

Cc≥Ch

Ch≥Cc

MER Values

Stream Splitting

• Two streams created from one

• one heat exchanger on each piece of split stream with couplings

1

1a

1b

1b

1a

1

Optimization of HEN

• How does approach delta T (ΔTmin) effect the total cost of HEN?

• Q= UA ΔT

• LW=QToΔT/(T1T2)

– More Utility cost

Costs

• Heat Exchanger Purchase Cost– CP=K(Area)0.6

• Annual Cost– CA=im[ΣCp,i+ ΣCP,A,j]+sFs+(cw)Fcw

• im=return on investment• Fs= Annual Flow of Steam,

– $5.5/ston to $12.1/ston

• Fcw=Annual Flow of Cold Water– $0.013/ston

Change ΔTmin

CP=K(Area)0.6

Area=Q/(UF ΔTmin)

More Lost Work

LW=QToΔT/(T1T2)

Optimization of Process Flowsheets

Chapter 24

Terry A. Ring

CHEN 5353

Degrees of Freedom

• Over Specified Problem – Fitting Data– Nvariables>>Nequations

• Equally Specified Problem – Units in Flow sheet– Nvariables=Nequations

• Under Specified Problem– Optimization– Nvariables<<Nequations

Optimization

• Number of Decision Variables– ND=Nvariables-Nequations

• Objective Function is optimized with respect to ND

Variables– Minimize Cost– Maximize Investor Rate of Return

• Subject To Constraints– Equality Constraints

• Mole fractions add to 1

– Inequality Constraints• Reflux ratio is larger than Rmin

– Upper and Lower Bounds• Mole fraction is larger than zero and smaller than 1

PRACTICAL ASPECTS• Design variables, need to be identified and kept

free for manipulation by optimizer – e.g., in a distillation column, reflux ratio specification

and distillate flow specification are degrees of freedom, rather than their actual values themselves

• Design variables should be selected AFTER ensuring that the objective function is sensitive to their values– e.g., the capital cost of a given column may be

insensitive to the column feed temperature

• Do not use discrete-valued variables in gradient-based optimization as they lead to discontinuities in f(d)

Optimization

• Feasible Region– Unconstrained Optimization

• No constraints– Uni-modal– Multi-modal

– Constrained Optimization• Constraints

– Slack– Binding

v

v

v

N

i ii=1

i V

N

ij j i Ej=1

N

ij j i Ij=1

MinimizeJ x f xd

Subject to (s.t.) x 0,i 1, ,N

a x b,i 1, ,N

c x d,i 1, ,N

LINEAR PROGRAMING (LP)

equality constraints

inequality constraints

objective function

w.r.t. design variables The ND design variables, d, are adjusted to minimize f{x} while satisfying the constraints

EXAMPLE LP – GRAPHICAL SOLUTION

A refinery uses two crude oils, with yields as below.

Volumetric Yields Max. Production

Crude #1 Crude #2 (bbl/day)

Gasoline 70 31 6,000

Kerosene 6 9 2,400

Fuel Oil 24 60 12,000The profit on processing each crude is:

$2/bbl for Crude #1 and $1.4/bbl for Crude #2.

a) What is the optimum daily processing rate for each grade?

b) What is the optimum if 6,000 bbl/day of gasoline is needed?

EXAMPLE LP –SOLUTION (Cont’d)Step 1. Identify the variables. Let x1 and x2 be the daily production rates of Crude #1 and Crude #2.

Step 2. Select objective function. We need to maximizemaximize profit: 1 2J x 2.00x 1.40x

Step 3. Develop models for process and constraints. Only constraints on the three products are given:

Step 4. Simplification of model and objective function. Equality constraints are used to reduce the number of independent variables (ND = NV – NE). Here NE = 0.

1 2

1 2

1 2

0.70x 0.31x 6,000

0.06x 0.09x 2,400

0.24x 0.60x 12,000

EXAMPLE LP –SOLUTION (Cont’d)

Step 5. Compute optimum. a) Inequality constraints define feasible space.

1 20.70x 0.31x 6,000

1 20.06x 0.09x 2,400

1 20.24x 0.60x 12,000Feasible

Space

EXAMPLE LP –SOLUTION (Cont’d)Step 5. Compute optimum. b) Constant J contours are positioned to find

optimum.

J = 10,000

J = 20,000

J = 27,097

x1 = 0, x2 = 19,355 bbl/day

EXAMPLE LP – GRAPHICAL SOLUTION

A refinery uses two crude oils, with yields as below.

Volumetric Yields Max. Production

Crude #1 Crude #2 (bbl/day)

Gasoline 70 31 6,000

Kerosene 6 9 2,400

Fuel Oil 24 60 12,000The profit on processing each crude is:

$2/bbl for Crude #1 and $1.4/bbl for Crude #2.

a) What is the optimum daily processing rate for each grade?

b) What is the optimum if 6,000 bbl/day of gasoline is needed?

Dealing with Impurities in Processes and Process Simulators

ChEN 5253 Design IITerry A. Ring

There is not chapter in the book on this subject

Impurity Effects

• Heat Exchange

• Reactors

• Separation Systems

• Recycle Loops

Impurities in Heat Exchange

• Impurities effect heat capacity– Lower Cp

• Various options

– Raise Cp

• Increase H2

• Impurities effect the enthalpy of stream– Total heat of condensation is less due to

impurity– Total heat of vaporization is less due to

impurity

Impurities in Separation Trains

• Non-condensable Impurities– Build up in Distillation column – Big Trouble!!

• Condensable Impurities– Cause some products to be less pure

• May not meet product specifications• Can not sell this product – Big Trouble!!

– Rework cost– Waste it– Sell for lower price

Processes are tested for Impurity Tolerance

• Add light and heavy impurities to feed– Low concentration

• All impurities add to 0.1 % of feed• (may need to increase Tolerance in Simulation)

– Medium concentration• All impurities add to 1% of feed

– High concentration• All impurities add to 10% of feed

• Find out where impurities end up in process• Find out if process falls apart due to impurities

– What purges are required to return process to function.

Impurities in Separation Trains

• It is important to know where the impurites will accumulate in the train

• Which products will be polluted by which impurities– Is that acceptable for sale of product?

Purging Impurities

• Find the point in the process where the impurities have the highest concentration– Put Purge here

• Put a purge in almost all recycle loops

Plant-Wide Controllability

• Control Architecture– DoF analysis Dynamic

Analysis• No. of valves

– DoF analysis Steady State Analysis

• No. of valves – No. of liquid level loops

• Product Flow Control or Feed Flow Control

• Types of control– Single loop PID– Gain Scheduling– Ratio control– Cascade Control– Multi-variable control– Model Based control

(MPC)– Override control

Distillation Control

• Types of Control– LV control– DV– LB– DB– (L/D)(V/B)– (L/F)(V/F)

The End